【HDU 5399】Too Simple

题

Description

Rhason Cheung had a simple problem, and asked Teacher Mai for help. But Teacher Mai thought this problem was too simple, sometimes naive. So she ask you for help.
Teacher Mai has m functions f1,f2,...,fm:{1,2,...,n}→{1,2,...,n}(that means for all x∈{1,2,...,n}，f(x)∈{1,2,...,n}.
But Rhason only knows some of these functions, and others are unknown.
She wants to know how many different function series f1,f2,...,fm there are that for every i(i≤1≤n),f1(f2(...(fm(i))...))=i. Two function series f1,f2,...,fm and g1,g2,...,gm are considered different if and only if there exist i(1≤i≤m), j(1≤j≤n),fi(j)≠gi(j)

Input

For each test case, the first lines contains two numbers n,m(1≤n,m≤100)The following are m lines. In i-th line, there is one number -1;or n space-separated numbers.

If there is only one number -1, the function fi is unknown. Otherwise the j-th number in the i-th line means fi(j)

Output

For each test case print the answer modulo 109+7.

Sample Input

3 3

1 2 3
-1

3 2 1

Sample Output

1

Hint

The order in the function series is determined. What she can do is to assign the values to the unknown functions.

题意：

求满足f1(f2(...(fm(i))...))=i的未知的函数有多少种可能。

分析：

因为只要最后的那个fi确定了，其它的任意都可以，答案是(n!)^(m-1)再mod 109+7,m为-1的个数

但是要注意只有一个-1时，答案可能是0或1，而当f里1..n没有全部出现时，即有重复数字时，答案也是0。

这题说是too simple，然而好多坑啊！样例只有一组数据，但是实际上可能有多组数据，除此，要注意每次处理新的一组时，哪些变量要清零，还有这题要用long long，n阶乘可以在一开始初始化。

代码：

#include<stdio.h>
#define M 1000000007LL
#define ll long long
#define N 105
#define F(a,b,c) for(int a=b;a<=c;a++)
ll n,m,d,f

,y
,jc
={1,1},ans;
int main()
{
F(i,2,100)jc[i]=jc[i-1]*i%M;//初始化阶乘
while(~scanf("%lld%lld",&n,&m))
{
d=0;ans=1;//初始化
F(i,1,m)
{
scanf("%lld",&f[i][1]);
if(f[i][1]==-1)d++;
else F(j,2,n)
{
scanf("%lld",&f[i][j]);
if(ans)F(k,1,j-1)
if(f[i][j]==f[i][k])ans=0;
}
}
if(ans)
{
if(d==0)
{
F(i,1,n)y[i]=i;
for(int i=m; i; i--)
F(j,1,n)y[j]=f[i][y[j]];//一层层推到f1
F(i,1,n&&ans)if(y[i]!=i)ans=0;
}
else
F(i,1,d-1)ans=ans*jc
%M;
}
printf("%lld\n",ans);
}
return 0;
}